$\overline{AB} = 3\sqrt{10}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $3\sqrt{10}$ $?$ $ \sin( \angle BAC ) = \frac{ \sqrt{10}}{10}, \cos( \angle BAC ) = \frac{3\sqrt{10} }{10}, \tan( \angle BAC ) = \dfrac{1}{3}$
Explanation: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{3\sqrt{10}} $ $ \overline{AC}=3\sqrt{10} \cdot \cos( \angle BAC ) = 3\sqrt{10} \cdot \frac{3\sqrt{10} }{10} = 9$